A candidate is required to answer $6$ out of $10$ questions which is divided into two groups $A$ and $B$ each containing $5$ questions. He is not permitted to attempt more than $4$ questions from either group. Find the number of different ways in which the candidate can choose six questions.

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Solution

Solution:
Number of choosing 6 questions from 10 $=^{10} C_{6} = 210.$
Number of attempting more than $4$ questions from the group $= 2 \times^{5} C_{5} \times^{5} C_{1} = 10.$
So, required number of ways $= 210 - 10 = 200.$
Alternative,
Reuired number of ways $=^{5} C_{3} \times^{5} C_{3} +^{5} C_{4} \times^{5} C_{2} +^{5} C_{2} \times^{5} C_{4} .$
$= 10 \times 10 + 5 \times 10 + 10 \times 5$
$= 100 + 50 + 50$
$= 200.$

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